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Let's consider a Brownian Motion $B$ defined on a probability space $(\Omega,\mathcal A,P)$; take $\mu>0$ and define the supermartingale $X_t:=B_t-\mu t$.

Pick $a,b>0$ and define the stopping time $\tau:=\inf\{s>0\;:\;X_s\notin]-a,b[\}$.

My book says that in order to prove that $\tau\in L^1(\Omega,\mathcal A,P)$ it's sufficient to prove that $$ \int_0^{+\infty}P(\tau>t)\,dt<+\infty $$ but I can't understand why.

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    Because for any nonnegative random variable $Z$, whether integrable or not, $$Z=\int_0^\infty\mathbf 1_{Z>t}\,dt$$ hence $$E(Z)=E\left(\int_0^\infty\mathbf 1_{Z>t}\,dt\right)=\int_0^\infty P(Z>t)\,dt$$ as has been explained tons of times on the site. – 2017-01-08

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